如何通过仅知道其基指针来删除整个对象

Question

我正在尝试用 C++ 实现自动释放池，但在解除分配时遇到了问题。所以我们有 root-class object:

class object {
 public:
    object() {
        retainCount_ = 0;
    }
    ~object() {
        std::cout << "object " << Description() << " is dealocated" << std::endl;
    }
    /* code with retain/release here */
    int retainCount() {
        return retainCount_;
    }
    std::string Description() {
        std::stringstream ss;
        ss << this;
        return ss.str();
    }
 private:
    int retainCount_;
};

还有一些实现：

class Integer : public object {
 public:
    int i;
    Integer(int ii) : i(ii) {}
    ~Integer() {
        std::cout << "Integer " << Description() << " is dealocated" << std::endl;
    }
};

当然还有释放池 class，它与 root-class 指针一起工作：

class release_pool {
 public:
    void insert(object* obj) {
        pointers_.insert(obj);
    }
    void flush() {
        std::set<object*>::iterator it = pointers_.begin();
        std::set<object*>::iterator tmp;
        const int N = pointers_.size();
        for (int i = 0; i < N; i++) {
            tmp = it;
            it++;
            if ((*tmp)->retainCount() == 0 ) {
                object* obj = *tmp;
                std::cout << "delete obj: " << obj->Description() << std::endl;
                pointers_.erase(tmp);
                delete obj;
            }
        }
    }
 private:
    std::set<object*> pointers_;
};

main.cpp 测试代码：

int main () {
    release_pool pool;

    Integer* obj = new Integer(5);
    pool.insert(obj);
    std::cout << "before flush: " << obj->i << "\n";

    pool.flush();

    std::cout << "after flush: " << obj->i << "\n";    
    return 0;
}

构建后，我得到下一个：

before flush: 5
delete obj: 0x7f9a84c025d0
object 0x7f9a84c025d0 is dealocated
after flush: 5

最后：调用 root-class 的析构函数，但不调用 Integer 的析构函数。因此，我们泄漏了分配给 Integer 对象的内存。你有什么想法来解决它吗？我如何删除整个对象，而不是它的根部分。

Answer 1

您需要将对象析构函数设为虚拟。

如何通过仅知道其基指针来删除整个对象

How to delete whole object by knowing only its base-pointer

c++

inheritance

release-management